Solving Equations with Technology: exam review Exam review Worked solutions

Three equations are \(2x + y = 3z + 11\), \(x - 2y + z = -2\) and \(4x + 5 = y - 2z + 13\). [5]

1. (a) Write the equations in the form \(ax + by + cz = d\). [2]

   • A1 \(2x + y - 3z = 11\)
   • A1 \(4x - y + 2z = 8\) (the second equation is already in this form)

2. (b) Use your calculator to find the values of \(x\), \(y\) and \(z\). [3]

   • M1 for entering three equations in three unknowns in the simultaneous equation solver
   • A1 for two correct values
   • A1 \(x = 3,\ y = 2,\ z = −1\)

Use your calculator in this question. Give answers to 3 significant figures. [6]

1. (a) Show that \(2x + \dfrac{5}{x} = 9\) can be written as \(2x^2 - 9x + 5 = 0\). [2]

   • M1 for multiplying every term by \(x\): \(2x^2 + 5 = 9x\)
   • A1 for moving the right-hand side across: \(2x^2 - 9x + 5 = 0\)

2. (b) Hence find the possible values of \(x\). [2]

   • M1 for entering the coefficients 2, −9, 5 in the polynomial solver
   • A1 \(x = 0.649\) or \(x = 3.85\)

3. (c) Find the real solution of \(3x^3 + 2x + 11 = 0\). [2]

   • M1 for entering the coefficients 3, 0, 2, 11 (0 for the missing \(x^2\) term)
   • A1 \(x = −1.40\) (the other two solutions are complex)

Solve each equation. Give real solutions to 3 significant figures. [6]

1. (a) Solve \(9x^2 - 12x + 4 = 0\). [2]

   • M1 for entering the coefficients 9, −12, 4
   • A1 \(x = 0.667\) (the solver gives one value: the root is repeated)

2. (b) Show that \(x^2 - 4x + 9 = 0\) has no real solutions. [2]

   • M1 for entering the coefficients 1, −4, 9
   • R1 the solutions are \(2 \pm 2.24i\): complex, so there are no real solutions

3. (c) Find the real solutions of \(x^4 - 5x^2 + x + 2 = 0\). [2]

   • M1 for entering the coefficients 1, 0, −5, 1, 2 (0 for the missing \(x^3\) term)
   • A1 \(x = −2.25, −0.555, 0.802, 2\)

The graph shows \(y = f(x)\), where \(f(x) = x^3 - 3x^2 - 9x\). Use your calculator. [6]

1. (a) Find the \(x\)-intercepts of the graph. [2]

   • M1 for solving \(f(x) = 0\) (polynomial solver, coefficients 1, −3, −9, 0) or reading the intercepts from the graph
   • A1 \(x = −1.85, 0, 4.85\)

2. (b) Solve \(f(x) = −10\). [2]

   • M1 for solving \(x^3 - 3x^2 - 9x = −10\), or for the intersections of the curve with the line \(y = −10\)
   • A1 \(x = −2.42, 0.917, 4.51\)

3. (c) The line \(y = c\) meets the curve at exactly two points. Find the possible values of \(c\). [2]

   • M1 for the line touching the curve at a turning point: the calculator gives a maximum of height \(5\) and a minimum of height \(−27\)
   • A1 \(c = −27\) or \(c = 5\)

Consider the equations \(3x - 2y = 5\) and \(6x - 4y = k\), where \(k\) is a constant. [4]

1. (a) Find the value of \(k\) for which the system has infinitely many solutions. [2]

   • M1 for the second equation being 2 times the first, so \(k = 2 \times 5\)
   • A1 \(k = 10\)

2. (b) Explain, using the graphs of the two equations, why the system has no solution when \(k = 12\). [2]

   • R1 both lines have gradient \(\frac{3}{2}\), so they are parallel
   • R1 for \(k = 12\) the lines are different (not the same line), so they never meet and there is no solution

Three numbers \(x\), \(y\) and \(z\) satisfy \(x \le y \le z\). Their mean is 8, their median is 7 and their range is 9. [5]

1. (a) Write down three equations in \(x\), \(y\) and \(z\). [3]

   • A1 \(x + y + z = 24\) (mean 8 of three numbers)
   • A1 \(y = 7\) (the middle value is the median)
   • A1 \(z - x = 9\)

2. (b) Find the three numbers. [2]

   • M1 for entering the three equations in the simultaneous equation solver
   • A1 \(x = 4,\ y = 7,\ z = 13\)

A curve has equation \(y = ax^2 + bx\). It passes through the point \((2,\ 6)\) and its gradient at that point is 7. [7]

1. (a) Write down two equations in \(a\) and \(b\). [2]

   • M1 for substituting \(x = 2,\ y = 6\): \(4a + 2b = 6\)
   • A1 \(4a + b = 7\) (from \(\frac{dy}{dx} = 2ax + b = 7\) at \(x = 2\))

2. (b) Find the values of \(a\) and \(b\). [2]

   • M1 for solving the two equations with the solver
   • A1 \(a = 2,\ b = −1\)

3. (c) The line \(y = 3x + 1\) meets the curve at the points P and Q. Use your calculator to find the distance PQ. [3]

   • M1 for \(2x^2 - x = 3x + 1\) or graphing the curve \(y = 2x^2 - x\) and the line
   • A1 \(x = −0.225\) and \(x = 2.22\) (so P and Q are \((−0.225,\ 0.326)\) and \((2.22,\ 7.67)\))
   • A1 PQ = 7.75

A company invests $24 000 in three funds for one year. Fund P pays 2% per year, fund Q pays 3.5% per year and fund R pays 5% per year. There is $4 000 more in fund Q than in fund P. The total interest earned over the year is $870. Let \(p\), \(q\) and \(r\) be the amounts, in dollars, in funds P, Q and R. [8]

1. (a) Write down an equation for the total amount invested. [1]

   • A1 \(p + q + r = 24000\)

2. (b) Write down an equation for the total interest earned. [2]

   • M1 for changing the percentages to decimals and multiplying by the amounts
   • A1 \(0.02p + 0.035q + 0.05r = 870\)

3. (c) Write down an equation using the fact that fund Q has $4 000 more than fund P. [1]

   • A1 \(-p + q = 4000\)

4. (d) Find the amount invested in fund R. [2]

   • M1 for entering the three equations in the simultaneous equation solver
   • A1 \(r = 8000\) in fund R: $8 000

5. (e) Find the percentage of the total interest that comes from fund R. Give your answer to 3 significant figures. [2]

   • M1 for \(\frac{0.05 \times 8000}{870} \times 100\) (interest from R is $400)
   • A1 46.0%

A craft stall makes candles, soaps and bath bombs. Each candle uses 300 g of wax and 10 ml of fragrance and sells for $15. Each soap uses 120 g of wax and 6 ml of fragrance and sells for $6. Each bath bomb uses 60 g of wax and 8 ml of fragrance and sells for $9. In one week the stall uses 10.8 kg of wax and 590 ml of fragrance, and everything made is sold for a total of $720. Let \(c\), \(s\) and \(b\) be the numbers of candles, soaps and bath bombs made. [7]

1. (a) Write down three equations in \(c\), \(s\) and \(b\). [3]

   • A1 \(300c + 120s + 60b = 10800\) (10.8 kg = 10 800 g)
   • A1 \(10c + 6s + 8b = 590\)
   • A1 \(15c + 6s + 9b = 720\)

2. (b) Find the number of bath bombs made. [2]

   • M1 for entering the three equations in the simultaneous equation solver
   • A1 \(b = 30\) (with \(c = 20\), \(s = 25\))

3. (c) Find the mean price at which an item is sold. Give your answer in dollars to 2 decimal places. [2]

   • M1 for \(\frac{720}{20 + 25 + 30}\)
   • (A1)(ft) $9.60

A rectangular sheet of metal measures 36 cm by 20 cm. A square of side \(x\) cm is cut from each corner and the sides are folded up to make an open tray. [8]

1. (a) Show that the volume of the tray is \(V = 4x^3 - 112x^2 + 720x\) cm³. [2]

   • M1 for length \(36 - 2x\), width \(20 - 2x\) and height \(x\): \(V = x(36 - 2x)(20 - 2x)\)
   • A1 expanding: \(V = 4x^3 - 112x^2 + 720x\)

2. (b) Explain why \(x\) must be less than 10. [1]

   • R1 the width \(20 - 2x\) must be positive, so \(x < 10\)

3. (c) The volume of the tray is 1000 cm³. Find the possible values of \(x\). [3]

   • M1 for \(4x^3 - 112x^2 + 720x - 1000 = 0\)
   • A1 the solver gives \(x = 1.93, 6.70, 19.4\)
   • A1 reject \(x = 19.4\) because it is greater than 10: \(x = 1.93\) or \(x = 6.70\)

4. (d) Find the length and the width of the tray when \(x\) takes the smaller possible value. [2]

   • A1 length \(36 - 2 \times 1.93 = 32.1\) cm
   • (A1)(ft) width \(20 - 2 \times 1.93 = 16.1\) cm

A charity records the money, in dollars, collected on days 1, 2, 3, … of a campaign. The amounts form an arithmetic sequence. On day 5 it collects $23. In the first 10 days it collects $250 in total. In a second campaign the amounts collected on days 1, 2 and 3 form a geometric sequence with first term 8. The total collected over these three days is $38. [10]

1. (a) Find the amount collected on day 1 and the common difference. [3]

   • M1 for \(u_1 + 4d = 23\)
   • M1 for \(\frac{10}{2}(2u_1 + 9d) = 250\), so \(2u_1 + 9d = 50\)
   • A1 \(u_1 = 7,\ d = 4\)

2. (b) The total amount collected in the first \(n\) days is $900. Find \(n\). [3]

   • M1 for \(\frac{n}{2}(2 \times 7 + (n - 1) \times 4) = 900\), which gives \(2n^2 + 5n - 900 = 0\)
   • A1 the solver gives \(n = −22.5\) or \(n = 20\)
   • A1 \(n = 20\) (a number of days is positive)

3. (c) In the second campaign, find the possible values of the common ratio \(r\). [3]

   • M1 for \(8 + 8r + 8r^2 = 38\), or for the sum formula with \(r \ne 1\)
   • A1 \(8r^2 + 8r - 30 = 0\)
   • A1 \(r = 1.5\) or \(r = −2.5\)

4. (d) State which value of \(r\) is possible for amounts of money collected, giving a reason. [1]

   • R1 \(r = 1.5\): with \(r = −2.5\) the amount on day 2 would be −$20, which is negative

A school hall is hired for a concert. The weekly income from \(x\) tickets sold is \(R = 80x - 0.5x^2\) dollars and the weekly cost is \(C = 1500 + 12x\) dollars. [8]

1. (a) Find the break-even quantities. [2]

   • M1 for graphing \(R\) and \(C\) (or solving \(80x - 0.5x^2 = 1500 + 12x\)) and finding the intersections
   • A1 \(x = 27.7\) and \(x = 108\) (108.3)

2. (b) Find the smallest and the largest whole number of tickets that give a profit. [2]

   • M1 for the profit lying between the break-even values
   • A1 28 and 108 (round into the profit range: 27.7 gives 28 and 108.3 gives 108)

3. (c) Find the greatest profit and the number of tickets that gives it. [2]

   • M1 for finding the maximum of \(P = 68x - 0.5x^2 - 1500\), or the greatest gap between the curve and the line
   • A1 $812 at \(x = 68\)

4. (d) The organisers want a profit of at least $600. Find the range of whole numbers of tickets that achieves this. [2]

   • M1 for the intersections of \(P\) with \(y = 600\): \(x = 47.4\) and \(x = 88.6\)
   • A1 48 to 88 tickets

A ball is thrown upwards from a platform. Its height above the ground after \(t\) seconds is \(h = −4.9t^2 + 14t + 2\) metres, for \(t \ge 0\). [7]

1. (a) Write down the height of the platform. [1]

   • A1 2 m (the value at \(t = 0\))

2. (b) Find the greatest height of the ball and the time at which it is reached. [2]

   • M1 for the maximum of the graph
   • A1 12 m at \(t = 1.43\) s

3. (c) Find the time at which the ball hits the ground. [2]

   • M1 for the positive root of \(h = 0\) (the other root, −0.136, is rejected)
   • A1 \(t = 2.99\) s

4. (d) Find the length of time for which the ball is higher than 6 m above the ground. [2]

   • M1 for the intersections of \(h\) with \(h = 6\): \(t = 0.322\) and \(t = 2.54\)
   • A1 \(2.535 − 0.322 = 2.21\) s